What is the solution for (dy)/(dx)=3xe^y,y(0)=0 ?

The solution for (dy)/(dx)=3xe^y,y(0)=0 is y=-ln(-(3x^2)/2+1)

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(dy)/(dx)=3xe^y,y(0)=0

\frac{d y}{d x} = 3 x e^{y}, y (0) = 0

y = - ln (- \frac{3 x ^{2}}{2} + 1)

Solution steps

\frac{d y}{d x} = 3 x e^{y}

Solve separable ODE:

y = - ln (- \frac{3 x ^{2}}{2} + 1)

y = - ln (- \frac{3 x ^{2}}{2} + 1)

\int_{1}^{7} \frac{2 ln ( x )}{x} d x

\int_{0}^{1} 6^{4 x} d x

y^{^{''}} - 3 y^{^{'}} = 18 t

\frac{d}{d t} (A e^{- 2 t} cos (t) + B e^{- 2 t} sin (t))

\frac{4 x}{x ^{2} + 2 x - 3}

What is the solution for (dy)/(dx)=3xe^y,y(0)=0 ?
The solution for (dy)/(dx)=3xe^y,y(0)=0 is y=-ln(-(3x^2)/2+1)